I have a few questions for Eliz. I am an after-school parapro who is trying to brush up on her math skills in order to better assist 8th and 9th grade students who are currently taking algebra or are trying to recover algebra credits.

One of the major topics students get entangled in is absolute value inequalities. I have found purple math's page on absolute value inequalities to be exceptionally helpful in helping students grasp the general rules/patters for various absolute value situations.

Q 1) Regarding |6t-2|<=6, I understand that you clear the absolute value bars by following the pattern for "less than" inequalities. I do not understand why you advise the next step to be

to divide though, on all three "sides", by 2. Then add 1 to all three "sides", and divide by 3.

Please explain why you would divide by 2 on all three "sides" when two is not attached to the variable? And why would you add 1 to all three sides?

I had assumed that, once the absolute value bars are cleared and the appropriate pattern is followed for the inequality, one would solve the middle "side" following the order of operations for equations, like so;

I admit that I have no idea if my answers are correct (I am thinking not, as they are not whole?) and am more interested in knowing the why behind your advised steps.

Q 2) In regards to |1-9x|=>3; again, I follow the the pattern for "greater than" inequalities, which requires you to split the inequality into two pieces, each representing the "positive" and "negative" directions the values are from zero. My question here about the split of the inequality is: does the piece with the negative sign always receive a flip of the inequality symbol?

|1-9x| => 3-----------------------1-9x < -3 or 1-9x > +3

Is it safe to make the direction of each inequality symbol in the negative and positive "pieces" of the inequality part of the rule or pattern when approaching absolute value inequalities?

Thanks so much in advance for your assistance in this matter! I am looking forward to having a much better grasp on these concepts and being able to successfully pass them on to my students!

anisaer wrote:Q 1) Regarding |6t-2|<=6, I understand that you clear the absolute value bars by following the pattern for "less than" inequalities. I do not understand why you advise the next step to be

to divide though, on all three "sides", by 2. Then add 1 to all three "sides", and divide by 3.

Please explain why you would divide by 2 on all three "sides" when two is not attached to the variable? And why would you add 1 to all three sides?

One doesn't "have" to do that step next but, since all the numbers involved are even, it certainly makes life simpler. Feel free to work with the larger numbers, if you prefer. You will want to work with fractions, though, rather than decimal approximations, if your students are to arrive at the correct answers.

anisaer wrote:Q 2) In regards to |1-9x|=>3; again, I follow the the pattern for "greater than" inequalities, which requires you to split the inequality into two pieces, each representing the "positive" and "negative" directions the values are from zero. My question here about the split of the inequality is: does the piece with the negative sign always receive a flip of the inequality symbol?

Don't you always flip the inequality symbol when you multiply through by a "minus"?

One doesn't "have" to do that step next but, since all the numbers involved are even, it certainly makes life simpler. Feel free to work with the larger numbers, if you prefer. You will want to work with fractions, though, rather than decimal approximations, if your students are to arrive at the correct answers.

Interesting. So you're advocating dividing all three sides by 2 as a way to reduce the solution to smaller numbers, like when reducing fractions, as opposed to following an order of operations process? I guess I didn't know that was "allowed", so to speak. I believe that in my students' case, typically solutions are in whole numbers, or the problem is given as a "which equation matches this solution set" dealeo.

Don't you always flip the inequality symbol when you multiply through by a "minus"?

Yes, you would flip the inequality sign when you divide or multiply by a negative. But I guess I don't see that as a step in the pattern. I understand splitting the equation into positive and negative sides to clear the absolute value bars because doing so represents values in opposing directions on a number line. But I did not associate that as being an equivalent action with multiplying or dividing by a negative, and thus necessitating a flip of the inequality sign. I am just wondering why that is an assumed part of the pattern as it's not really touched on in the article on absolute value inequalities; it is shown in the "greater than" example that the sign for the negative "piece" is flipped, but the only reasoning given is that "this is the pattern", not that it is the result of dividing or multiplying by a negative.